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Re: newbie is looking for a customDistribution function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg50402] Re: newbie is looking for a customDistribution function
*From*: Paul Abbott <paul at physics.uwa.edu.au>
*Date*: Thu, 2 Sep 2004 04:34:27 -0400 (EDT)
*Organization*: The University of Western Australia
*References*: <ch3o86$t96$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
In article <ch3o86$t96$1 at smc.vnet.net>, János <janos.lobb at yale.edu>
wrote:
> I looked for it in the archives, but found none.
It is there at
http://groups.google.com/groups?threadm=6d0ch7%242no%40smc.vnet.net
Also see The Mathematica Journal 1(3): 57, which is referenced at this
link. Further comments are given below.
> I am looking for ways
> to create a custom distribution, which I can call as a function. Here
> is an example for illustration. Let's say I have a list created from a
> 4 elements alphabet {a,b,c,d}:
>
> In[1]:=
> lst={a,a,b,c,a,d,a,c,c,a}
>
> Out[1]=
> {a,a,b,c,a,d,a,c,c,a}
>
> Distribute gives me - thanks David Park - all the two element
> combinations of {a,b,c,d}
>
> In[11]:=
> twocombs=Distribute[Table[{a,b,c,d},{2}],List]
>
> Out[11]=
> {{a,a},{a,b},{a,c},{a,d},{b,a},{b,b},{b,c},{b,d},{c,a},{c,b},{c,c},{c,d}
> ,{
> d,a},{d,b},{d,c},{d,d}}
>
> I can count the occurrence of an element of twocombs in lst with the
> following function:
>
> occuranceCount[x_List] := Count[Partition[lst, 2, 1], x]
>
> Mapping this function over twocombs gives me the number of occurances
> of elements of twocombs in lst:
>
> In[12]:=
> distro=Map[occuranceCount,twocombs]
>
> Out[12]=
> {1,1,1,1,0,0,1,0,2,0,1,0,1,0,0,0}
>
> It shows that for example {c,a} occurs twice, {d,a} occurs once and
> {d,c} or {d,d} never occur.
>
> Now, I would like to create a distribution function called
> twocombsLstDistribution which I could call and it would give me back
> elements of twocombs with the probability as they occur in distro, that
> is for on average I would get twice as much {c,a}s as {d,a}s and never
> get {d.c} or {d,d}.
>
> How can I craft that ?
The idea of the code below is to count for how many symbols the
cumulative frequencies
cumfreq[x_List] := FoldList[Plus, First[x], Rest[x]]/Tr[x];
are less than a fixed random number t in the range [0,1], and use the
number of hits as the index into the alphabet.
index[f_, r_] := Length[Select[f, r >= #1 & ]] + 1;
rand[x_List, cf_List] := x[[index[cf, Random[]]]]
For your distribution,
cf = cumfreq[distro]
here is a randome set of elements in twocombs with the probability as
they occur in distro.
Table[rand[twocombs, cf], {2000}];
As a check we see that
Count[%, #] & /@ twocombs
looks fine.
> /Of course I need it for an arbitrary but finite length string lst over
> a fixed length alphabet {a,b,c,d,....} for k-length elements of kcombs,
> and it has to be super fast :). My real lst is between 30,000 and
> 70,000 element long over a four element alphabet and I am looking for k
> between 5 and a few hundred. /
Indexing using zeroth-order Interpolation is considerably faster (See
e.g., http://groups.google.com/groups?selm=b34q2o%24gc1%241%40smc.vnet.net):
int[distro_] := int[distro] = Interpolation[Transpose[
{
Range[0, 1, 1/Tr[distro]],
Join[{1}, Flatten[MapIndexed[Table[First[#2], {#1}] & , distro]]]
}
], InterpolationOrder -> 0]
If you compare
SeedRandom[1];
Timing[test1 = Table[rand[twocombs, cf], {100000}];]
to
SeedRandom[1];
Timing[test2 =Table[twocombs[[int[distro][Random[]]]],{100000}];]
you should find that test1 == test2 and that using int[distro] is about
4 times faster.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 9380 2734
School of Physics, M013 Fax: +61 8 9380 1014
The University of Western Australia (CRICOS Provider No 00126G)
35 Stirling Highway
Crawley WA 6009 mailto:paul at physics.uwa.edu.au
AUSTRALIA http://physics.uwa.edu.au/~paul
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